Symmetry of a tensor of type $(0,4)$ Let $A$ be a tensor of type $(0,4)$ which satisfies the following symmetries:
\begin{align*}
A_{ijkl} &= - A_{jikl}, \\
A_{ijkl} &= - A_{ijlk}, \\
A_{ijkl} + A_{iklj} + A_{iljk} &= 0, \\
A_{ijkl} &= A_{klij}.
\end{align*}
How can I prove that if $A(v,w,v,w)= 0$ for any $v,w\in V$, then $A = 0 ?$
 A: Since $A$ has antisymmetry in the first pair and the last pair of arguments, it can be understood as a bilinear form acting on bivectors $v \wedge w \in \bigwedge^2 V$: i.e. 
$$b(x \wedge y,z \wedge w) := A(x,y,z,w) $$
is well-defined. Since $A$ is also symmetric, $b$ is a symmetric bilinear form. If we write $K(u \wedge v) = A(u,v,u,v)$, then $K(u \wedge v)=b(u\wedge v, u \wedge v)$, and so it follows that $K$ is a quadratic form on $\bigwedge^2 V$. The polarisation identity implies that a quadratic form corresponds to a unique bilinear form, so in fact we can go the other way: $K$ determines $b$ (and thus $A$) uniquely. Hence if $K=0$, $A=0$.

To get an explicit formula for $A$ in terms of $k(u,v)=A(u,v,u,v)$, one can first use linearity to express $A(x,y,z,w)$ in terms of things of the form $A(x,y,x,w)$:
$$ A(x+z,y,x+z,w) = A(x,y,x,w)+A(z,y,z,w)+A(x,y,z,w)+A(z,y,x,w) \\
-A(y+z,x,y+z,w) = -A(y,x,y,w)-A(z,x,z,w)+A(x,y,z,w)+A(x,z,y,w) \\
0 = A(x,y,z,w) + A(y,x,z,w) $$
Adding,
\begin{align}
&A(x+z,y,x+z,w)-A(y+z,x,y+z,w) \\
&= 3A(x,y,z,w) + A(x,y,x,w)+A(z,y,z,w)-A(y,x,y,w)-A(z,x,z,w) \\
&\quad + \big( A(z,y,x,w) + A(x,z,y,w) + A(y,x,z,w) \big)
\end{align}
The term in brackets vanishes by the Bianchi-type identity of antisymmetry in the last three arguments, so we have
$$ 3A(x,y,z,w) = A(x+z,y,x+z,w)-A(y+z,x,y+z,w) -  A(x,y,x,w) - A(z,y,z,w) +A(y,x,y,w) + A(z,x,z,w). \tag{*}  $$
Now, $A(x,\cdot,x,\cdot)$ is a symmetric bilinear form in $y,w$, with corresponding quadratic form $k(x,\cdot)$ so the ordinary polarisation identity lets us write
$$ 2A(x,y,x,w) = k(x,y+w) - k(x,y)-k(x,w). $$
Substituting this into (*) gives a (lengthy) formula for $A(x,y,z,w)$ in terms of $k$.
(The above adapted from Kühnel's Differential Geometry, Theorem 6.5)
